What is the solution for y^{''}+2y+8=0 ?

The solution for y^{''}+2y+8=0 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-4

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y^{''}+2y+8=0

y^{^{''}} + 2 y + 8 = 0

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 4

Solution steps

y^{''} (t) + 2 y + 8 = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 4

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 4

ln (sec (ln (x)))

\int - \frac{1}{u ^{2}} d u

\frac{\partial}{\partial y} (\frac{- y ^{2}}{( x - y ) ^{2}})

\int (x)^{2} d x

\int \frac{1}{x ln ( x ) ( ln ( ln ( x ) ) ) ^{2}} d x

What is the solution for y^{''}+2y+8=0 ?
The solution for y^{''}+2y+8=0 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-4